1CHAPTER 1 ExercisesE1.1 Charge = Current ×Time = (2 A) ×(10 s) = 20 C E1.2 A)2cos(200)200cos(2000.010t)0.01sin(20()()(ttdtddttdqti=×===E1.3Because i2has a positive value, positive charge moves in the same direction as the reference. Thus, positive charge moves downward in element C. Because i3has a negative value, positive charge moves in the opposite direction to the reference. Thus positive charge moves upward in element E. E1.4 Energy = Charge ×Voltage = (2 C) ×(20 V) = 40 J Because vabis positive, the positive terminal is aand the negative terminal is b. Thus the charge moves from the negative terminal to the positive terminal, and energy is removed from the circuit element.E1.5 iabenters terminal a. Furthermore, vabis positive at terminal a. Thus the current enters the positive reference, and we have the passive reference configuration. E1.6 (a) 220)()()(ttitvtpaaa==J666732032020)(310031001002=====∫∫ttdttdttpwaa(b) Notice that the references are opposite to the passive sign convention. Thus we have: 20020)()()(−=−=ttitvtpbbbJ100020010)20020()(1002100100−=−=−==∫∫ttdttdttpwbb
2E1.7 (a) Sum of currents leaving = Sum of currents entering ia= 1 + 3 = 4 A (b) 2 = 1 + 3 + ib⇒ib= -2 A (c) 0 = 1 + ic+ 4 + 3 ⇒ic= -8 A E1.8 Elements Aand Bare in series. Also, elements E, F, and G are in series.E1.9 Go clockwise around the loop consisting of elements A, B, and C: -3 - 5 +vc= 0 ⇒vc= 8 V Then go clockwise around the loop composed of elements C, Dand E: -vc- (-10) + ve= 0 ⇒ve= -2 V E1.10 Elements Eand Fare in parallel; elements Aand Bare in series. E1.11 The resistance of a wire is given by ALRρ=. Using 4/2dAπ=and substituting values, we have: 4/)106.1(1012.16.9236−−×××=πL⇒L= 17.2 m E1.12 RVP2=⇒Ω==144/2PVR⇒A833.0144/120/===RVIE1.13 RVP2=⇒V8.15100025.0=×==PRVmA8.151000/8.15/===RVIE1.14 Using KCL at the top node of the circuit, we have i1= i2. Then, using KVL going clockwise, we have -v1- v2= 0; but v1= 25 V, so we have v2 = -25 V. Next we have i1= i2= v2/R= -1 A. Finally, we have W25)1()25(22=−×−==ivPRand W.25)1()25(11−=−×==ivPsE1.15 At the top node we have iR= is= 2A. By Ohm’s law we have vR= RiR= 80 V. By KVL we have vs= vR= 80 V. Then ps= -vsis = -160 W (the minus sign is due to the fact that the references for vsand isare opposite to the passive sign configuration). Also we have W.160==RRRivP
3Answers for Selected ProblemsP1.7* Electrons are moving in the reference direction (i.e., from ato b). C9=QP1.9*( )A22tti+=P1.12* coulombs2=QP1.14* (a) km6.17=h(b) m/s9.587=v(c) The energy density of the battery is J/kg108.1723×which is about 0.384% of the energy density of gasoline. P1.17* coulombs106.35×=Qjoules10536.4Energy6×=P1.20* (a) 30 W absorbed (b) 30 W absorbed (c) 60 W supplied P1.22* C50=Q. Electrons move from b to a. P1.24* kWh500Energy=W694.4=PA787.5=I%64.8Reduction=P1.27* (a) P=50 W taken from element A.